Question

Find the distance between the parallel planes $\overrightarrow{r}.(2\hat{i}-3\hat{j}+6\hat{k})=5$ and $\overrightarrow{r}.(6\hat{i}-9\hat{j}+18\hat{k})+30=0$.

Answer 1

Plane 1 : $\overrightarrow{r}.(2\hat{i}-3\hat{j}+6\hat{k})=5$

Cartesian form : $2x-3y+6z-5=0-\left(1\right)$

Plane 2 : $\overrightarrow{r}.(6\hat{i}-9\hat{j}+18\hat{k})+20=0$

Cartesian form : $6x-9y+18z+20=0-\left(2\right)$

Taking plane (1)

putting $y,z=0$

we get $x=\frac{5}{2}$

Co-ordinates on plan 1 = $(\frac{5}{2},0,0)$

Distance of $(\frac{5}{2},0,0)$ from plane $6x-9y+18z+20=0$

$\Rightarrow \frac{|6\times \frac{5}{2}+(-9\left)\right(0)+18(0)+20|}{\sqrt{(6{)}^2+(9{)}^2+(18{)}^2}}$

$\Rightarrow \frac{35}{21}=\frac{5}{3}$ units